The above version of the \(\sqrt{3}\) ladder is standard, but finding a general term for the sequences \(\{a_{n}\}\) and \(\{b_{n}\}\) from the rungs \(\langle a_{n}\quad b_{n}\rangle\) as presented is not immediate, since the rungs have been reduced when possible. Here is the \(\sqrt{3}\) Greek Ladder without rung reductions: \[ \begin{array}{cc} 1 & 1\\ 2 & 4\\ 6 & 10\\ 16 & 28 \end{array} \qquad \begin{array}{cc} 44 & 76\\ 120 & 208\\ 328 & 568\\ 896 & 1552 \end{array} \] And now, a recursive definition of the rungs \(\langle a_{n}\quad b_{n}\rangle\) of this equivalent version of the \(\sqrt{3}\) Greek Ladder is suggested and given by \[ \begin{align*} a_{1} & =1\text{, }a_{2}=2\text{, and for }n>2\text{, }a_{n}=2a_{n-1} +2a_{n-2};\\ b_{1} & =1\text{, }b_{2}=4\text{, and for }n>2\text{, }b_{n}=2b_{n-1}+2b_{n-2} \end{align*} \] (It is left to the reader to show that this is equivalent to the standard definition given earlier for the \(\sqrt{3}\) ladder.) Using the same method of [4] again, the \(n\)-th rung \(\langle a_{n}\quad b_{n}\rangle\) is \[ \left\langle \frac{\left( 1+\sqrt{3}\right) ^{n}-\left( 1-\sqrt{3}\right)^{n}}{2\sqrt{3}}\qquad\frac{\left( 1+\sqrt{3}\right) ^{n}+\left( 1-\sqrt{3}\right) ^{n}}{2}\right\rangle\]

So let us see what happens when \(\frac{b_{n}}{a_{n}}\) is chosen as an estimate in Newton's Method. The corresponding approximation for \(\sqrt{3}\) deriving from the \(n\)-th rung simplifies to \[ \frac{b_{n}}{a_{n}} = \frac{\sqrt{3}\left( \sqrt{3}+1\right) ^{n}+\sqrt{3}\left( 1-\sqrt{3}\right) ^{n}}{\left( \sqrt{3}+1\right) ^{n}-\left( 1-\sqrt{3}\right) ^{n}} \] When this fraction is subjected to Newton's Method for \(\sqrt{3}\), you can check that the result is \[ \frac{\sqrt{3}\left[ \left( 1+\sqrt{3}\right) ^{2n}+\left( 1-\sqrt{3}\right) ^{2n}\right] }{\left(1+\sqrt{3}\right) ^{2n}-\left(1-\sqrt{3}\right) ^{2n}} \] which is the reduced form of the fraction \(\frac{b_{2n}}{a_{2n}}\) from the \(2n\)-th rung of the ladder, so the doubling is established for the \(\sqrt{3}\) ladder.